Probability exponential random variable smallest among others. For $X_1,\dots,X_n$ exponential random variables with mean $E(X_i)=\mu_i$. Now I want to calculate the probability that $X_i$ is the smallest among $X_1,\dots,X_n$. Therefore I am trying to calculate the $P(X_i=\min({X_1,\dots,X_n}))$. 
Now I have already calculated the distribution of the random variable $Y_n=\min({X_1,\dots,X_n})$ which can be expressed by $P(Y_n\leq t)=1-(e^{-\mu t})^n$.
I assume I need to use this with conditional probability in some way, but I do not exactly know how. Anyone has suggestions? thx
 A: Fix some $i$ and note that, for every $x$ and every $k$, $$P(X_k>x)=e^{-\lambda_kx}$$ where $$\lambda_k=\frac1{\mu_k}$$ hence $$P\left(\min_{k\ne i}X_k>x\right)=P\left(\bigcap_{k\ne i}\{X_k>x\}\right)=\prod_{k\ne i}e^{-\lambda_kx}=e^{-(\lambda-\lambda_i)x}$$ where $$\lambda=\sum_i\lambda_i$$ Since $X_i$ is independent of $(X_k)_{k\ne i}$, this implies
$$P\left(\min_{k\ne i}X_k>X_i\,{\large\mid}\, X_i\right)=e^{-(\lambda-\lambda_i)X_i}$$ Integrating both sides with respect to the distribution of $X_i$, one gets $$P\left(\min_{k\ne i}X_k>X_i\right)=E\left(e^{-(\lambda-\lambda_i)X_i}\right)=\int_0^\infty e^{-(\lambda-\lambda_i)x}\lambda_ie^{-\lambda_ix}dx$$ and finally, for every $i$,

$$P\left(X_i=\min_kX_k\right)=\frac{\lambda_i}\lambda$$

A: If you have a sequence of independent and identically distributed continuous random variables, then the index of the smallest instance among them will be discretely uniformly distributed.   Because any one of them may turn out to be the smallest among them with no bias.
$$\begin{align}\forall j\in\{2,..,n\}\,&~\Bigl[\,\mathsf P(X_1{=}\min_{i=1}^n\{X_i\})~=~\mathsf P(X_j{=}\min_{i=1}^n\{X_i\})\,\Bigr]\\[2ex]\sum_{j=1}^n \mathsf P(X_j{=}\min_{i=1}^n\{X_i\})~&=~1\\[2ex]\hline\therefore~\mathsf P(X_1{=}\min_{i=1}^n\{X_i\})~&=~1/n\end{align}$$
